Question

Stiind ca s-au amestecat cantitati egale de solutii de CuSO4 si NaOH si ca in final solutia obtinuta contine 25% sare si 10% baza , calculeaza concentratiile initiale ale solutiilor de CuSO4 si respectiv NaOH

Answer 1

·CuSO4 + 2NaOH ⇒ Na2SO4 + Cu(OH)2
ms1 = ms2 = m
ms final = 2m 
initial : n CuSO4 = (c1·m/100)/160      nNaOH =( c2·m/100)/40 
c2·m /40 =2·c1·m/160      c2 = c1/2
n Na2SO4 = n CuSO4 = nCu(OH)2 = [m·c1/100]/160  = nNaOH /2 = =[m·c2/100]/80
m sare = [m·c1/160]100 ·142 = 0,8875m·c1/100
[0,8875·m·c1]/100 / (2m) ·100 = 25      0,8875·c1 = 50       c1 = 56,338%
m baza ={ [m·c2/100] /80}·98
[(mc2/80)·98]/2m = 10     1,225c2= 20  c2 = 16/326%